Đáp án:
$\begin{array}{l}
b)\widehat A = {90^0};\widehat B = {30^0}\\
\Leftrightarrow \widehat C = {90^0} - {30^0} = {60^0}\\
+ \sin \widehat B = \dfrac{{AC}}{{BC}}\\
\Leftrightarrow \sin {30^0} = \dfrac{{10}}{{BC}} = \dfrac{1}{2}\\
\Leftrightarrow BC = 20\left( {cm} \right)\\
\Leftrightarrow AB = \sqrt {B{C^2} - A{C^2}} \left( {theo\,Pytago} \right)\\
= \sqrt {{{20}^2} - {{10}^2}} = 10\sqrt 3 \left( {cm} \right)\\
Vậy\,\widehat C = {60^0};BC = 20cm;AC = 10\sqrt 3 cm\\
f)\widehat B + \widehat C = {90^0};\widehat B = 2\widehat C\\
\Leftrightarrow 3\widehat C = {90^0}\\
\Leftrightarrow \widehat C = {30^0}\\
\Leftrightarrow \widehat B = {60^0}\\
\sin \widehat B = \dfrac{{AC}}{{BC}}\\
\Leftrightarrow \sin {60^0} = \dfrac{{\sqrt 3 }}{2} = \dfrac{{50}}{{BC}}\\
\Leftrightarrow BC = \dfrac{{100\sqrt 3 }}{3}\left( {cm} \right)\\
\Leftrightarrow AB = \sqrt {{{\left( {\dfrac{{100\sqrt 3 }}{3}} \right)}^2} - {{50}^2}} = \dfrac{{50\sqrt 3 }}{3}\left( {cm} \right)\\
Vậy\,\widehat B = {60^0};\widehat C = {30^0};BC = \dfrac{{100\sqrt 3 }}{3}cm;AB = \dfrac{{50\sqrt 3 }}{3}cm
\end{array}$
